Linear Independence of Two Functions Given the functions $2x$ and $|x|$ show that
   (a) these two vectors are linearly independent in $C[-1,1]$
   (b) the vectors are linearly dependent in $C[0,1]$.
My try:
Wronskian of $2x$, $|x|$ is identically zero, which gives no information whether the functions are linearly independent.
So for (a), I substitute $x$ for $-1$ one time and $1$ another one, in the linear combination of the two functions.
This produce the system: $$-2c1 + c2 = 0 $$ $$2c1 + c2 = 0$$
 The only solution of this system is $c1 = c2 = 0 $, and therefore the system is linearly independent.
For (b), the same procedure was conducted with $x = 1$, and $x = 1/2$, and the produced system is: $$2c1 + c2 = 0 $$ $$c1 + 1/2c2 = 0$$.
It's clear that the two equations are dependent on each other.
My question is: is my solution for (b) correct? the book, in the example, substitute $x$ with the boundary values, but for (b), $zero$ will eliminate the whole equation.
Also if we can put x to be any particular value within the given range, can we use this method before trying Wronskian? Sometimes it's faster and easier than computing derivatives.
 A: An alternative, perhaps easier, approach:
The functions$\;2x\;,\;\;|x|\;$ are linearly dependen somewhere iff one of them is a scalar multiple of the other one, so suppose
$$\exists\,k\in\Bbb R\;\;s.t.\;\;2xk=|x|=\begin{cases}\;\;\,x&,\;\;0\le x\le1\\{}\\-x&,\;\;-1\le x<0\end{cases}$$
But then we get
$$\begin{cases}x=\frac12\implies 2\cdot\frac12\cdot k=\left|\frac12\right|=\frac12\implies& \color{red}{k=\frac12}\\{}\\
x=-\frac12\implies 2\cdot\left(-\frac12\right)\cdot k=\left|-\frac12\right|=\frac12\implies& \color{red}{k=-\frac12}\end{cases}\;\implies\text{contradiction}$$
Part (b) is now, I think, clear: the functions on $\;[0,1]\;$ are $\;|x|=x\;,\;\;2x\;$ , and clearly the second one is twice the first one.
A: Your solution for b) is incorrect. Easiest is to notice that on $[0,1]$ , $|x|=x$, so your functions become $2x$ and $x$ which are obviously linearly dependent. Note that  $2$ vectors in a vector space are linearly dependent if and only if one is a scalar multiple of the other. 
